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yKpoI14uk [10]
3 years ago
8

In which situations is work being done? ( You can pick more than 1)

Physics
2 answers:
pickupchik [31]3 years ago
7 0
A balloon expanding would count as work because it is moving and none of the others are. However, I think that James pedaling also works because at least he is using his leg muscles even though he's not moving. He is working because he is exercising himself and that count as working. I would chose James over all of them.
Ivan3 years ago
4 0

Answer:

Andy’s car rolls down a hill with the engine off.

A balloon expands as it is heated by the Sun.

James pedals a stationary exercise cycle that doesn’t move.

Explanation:

1).  Hannah leans against her locker, but it doesn’t move.

Since Hannah is at rest and there is no displacement so there is no work on Hannah by the contact force

2).  Andy’s car rolls down a hill with the engine off.

Since car is rolled down the hill so here work is done on the car by force of gravity on the car

3). A wind turbine stands still on a sunny, windless day.

Since wind turbine is at rest so there will be no work on the turbine due to force of wind

4).  A balloon expands as it is heated by the Sun.

Here since the volume of balloon is increased so work is done by the internal air of balloon to expand it.

5).  James pedals a stationary exercise cycle that doesn’t move.

While pedaling since the force exerted by the cyclist and the pedal goes down so work is done by the cyclist to move the pedal up and down

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Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

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a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

Then, α = √3 at/2R, equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

α = √3 at/2R

α = √3√(a²-ar²)/2R

α = √3√(a²-(w²R)²)/2R

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Also, a = 2•ar = 2w²R

Then,

α = √3((2w²R)²-w⁴R²) / 2R

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3 years ago
A 1400 kg car driving at 25 m/s slams on its brakes. The coefficient of kinetic friction between the tires and the road is 0.7.
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The acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

The given parameters;

  • mass of the car, m = 1400 kg
  • Initial velocity of the car, u = 25 m/s
  • coefficient of kinetic friction, μ = 0.7

The acceleration of the car is calculated as follows;

a = μg

a = 0.7 x 9.8

a = 6.86 m/s²

The time taken for the car to stop is calculated by using Newton's second law of motion;

F = ma

F = \frac{mv}{t} \\\\ma = \frac{mv}{t}\\\\a = \frac{v}{t} \\\\t = \frac{v}{a} \\\\t = \frac{25}{6.86} \\\\t = 3.64 \ s

Thus, the acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

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